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Question Number 200302 by Calculusboy last updated on 16/Nov/23

Answered by Rasheed.Sindhi last updated on 18/Nov/23

(√(a+bx)) +(√(b+cx)) +(√(c+ax)) =(√(b−ax)) +(√(c−bx)) +(√(a−cx)) a+bx≥0 ∧ b+cx≥0 ∧ c+ax≥0 (a+b+c)+(a+b+c)x≥0⇒x≥−((a+b+c)/(a+b+c))=−1 determinant (((x≥−1))) b−ax≥0 ∧ c−bx≥0 ∧ a−cx≥0 (a+b+c)−(a+b+c)x≥0⇒(a+b+c)x≤a+b+c determinant (((x≤1))) determinant (((−1≤x≤1))) .....

a+bx+b+cx+c+ax=b−ax+c−bx+a−cxa+bx⩾0∧b+cx⩾0∧c+ax⩾0(a+b+c)+(a+b+c)x⩾0⇒x⩾−a+b+ca+b+c=−1x⩾−1b−ax⩾0∧c−bx⩾0∧a−cx⩾0(a+b+c)−(a+b+c)x⩾0⇒(a+b+c)x⩽a+b+cx⩽1−1⩽x⩽1.....

Commented by Calculusboy last updated on 20/Nov/23

thanks sir

thankssir

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